View source for 1995 AHSME Problems/Problem 27

==Problem==
Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.

<cmath>\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\
& & & & 1 & & 1 & & & & \\
& & & 2 & & 2 & & 2 & & & \\
& & 3 & & 4 & & 4 & & 3 & & \\
& 4 & & 7 & & 8 & & 7 & & 4 & \\
5 & & 11 & & 15 & & 15 & & 11 & & 5 & \end{tabular}</cmath>

Let <math>f(n)</math> denote the sum of the numbers in row <math>n</math>. What is the remainder when <math>f(100)</math> is divided by 100?

<math> \mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 }  </math>

==Solution==
We sum the first few rows: 0, 2, 6, 14, 30, 62. They are each two less than a power of 2, so we try to prove it:

Let the sum of row <math>n</math> be <math>S_n</math>. To generate the next row, we add consecutive numbers. So we double the row, subtract twice the end numbers, then add twice the end numbers and add two. That makes <math>S_{n+1}=2S_n-2(n-1)+2(n-1)+2=2S_n+2</math>. If <math>S_n</math> is two less than a power of 2, then it is in the form <math>2^x-2</math>. <math>S_{n+1}=2^{x+1}-4+2=2^{x+1}-2</math>. Since the first row is two less than a power of 2, all the rest are. Since the sum of the elements of row 1 is <math>2^1-2</math>, the sum of the numbers in row <math>n</math> is <math>2^n-2</math>. Thus, using [[Modular arithmetic]], <math>f(100)=2^{100}-2\equiv 2^{20}-2\bmod{100}</math>. <math>2^{10}=1024</math>, so <math>2^{20}-2\equiv 24^2-2\equiv 74\bmod{100} \Rightarrow \mathrm{(E)}</math>.

==See also==
{{AHSME box|year=1995|num-b=27|num-a=29}}

[[Category:Introductory Number Theory Problems]]